The figure (Figure 1) shows four sides of a 3.0 cm x 3.0 cm x 3.0 cm cube
 |
| Figure 1 |
-----------------------------------------------------------------------------------------------------
PART A:
What is the electric fluxe Φ1 through side 1?
Express your answer using two significant figures.
SOLUTION:
Begin by converting all dimensions to SI units, as per usual.
l = w = h = 0.03 m
Assign a coordinate system.
For this walkthrough, let the lower left corner of the cube be the origin. The x-axis is positive horizontally to the right and the y-axis is positive going upward vertically.
Calculate the surface area... this is very simple since the object is a cube. This means that any side, regardless of orientation, has the same surface area.
A = l⋅w = l⋅h = w⋅h
→ A = 9×10-4 m2
→ A = 9×10-4 m2
Next determine the electric field vector.
The magnitude of the electric field is ... | E | = 500 N/C
The unit vector: vx = cos(30)i^ & vy = sin(30)j^
V = cos(30)i^ + sin(30)j^
→ E = [500 N/C]⋅[cos(30)i^ + sin(30)j^]
→ E = [500 N/C]⋅[cos(30)i^ + sin(30)j^]
The significant surface in this part is side 1.
The area vector describing side 1 is...
→ A1 = (9×10-4 m2)i^
ΦE = ∫E⋅dA
Φ1 = [500 N/C]⋅[cos(30)i^ + sin(30)j^]⋅(9×10-4 m2)i^
→Φ1 = 0.3897 N⋅m2/C
Answer is negative. I'm not positive why, though.
→Φ1 = -0.39 N⋅m2/C
Answer is negative. I'm not positive why, though.
→Φ1 = -0.39 N⋅m2/C
-----------------------------------------------------------------------------------------------------
PART B:
What is the electric fluxe Φ2 through side 2?
Express your answer using two significant figures.
SOLUTION:
The beginning of this part is exactly like part A.
A = 9×10-4 m2
E = [500 N/C]⋅[cos(30)i^ + sin(30)j^]
E = [500 N/C]⋅[cos(30)i^ + sin(30)j^]
The significant surface in this part is side 2.
The area vector describing side 2 is...
→ A1 = (9×10-4 m2)j^
ΦE = ∫E⋅dA
Φ2 = [500 N/C]⋅[cos(30)i^ + sin(30)j^]⋅(9×10-4 m2)j^
→Φ2 = 0.225 N⋅m2/C
-----------------------------------------------------------------------------------------------------
PART C:
What is the electric fluxe Φ3 through side 3?
Express your answer using two significant figures.
Express your answer using two significant figures.
SOLUTION:
The beginning of this part is exactly like part A.
A = 9×10-4 m2
E = [500 N/C]⋅[cos(30)i^ + sin(30)j^]
The significant surface in this part is side 3.
The area vector describing side 3 is...
The area vector describing side 3 is...
→ A3 = (9×10-4 m2)i^
ΦE = ∫E⋅dA
Φ3 = [500 N/C]⋅[cos(30)i^ + sin(30)j^]⋅(9×10-4 m2)i^
→Φ3 = 0.39 N⋅m2/C
-----------------------------------------------------------------------------------------------------
PART D:
What is the electric fluxe Φ4 through side 4?
Express your answer using two significant figures.
Express your answer using two significant figures.
SOLUTION:
The beginning of this part is exactly like part A.
A = 9×10-4 m2
E = [500 N/C]⋅[cos(30)i^ + sin(30)j^]
E = [500 N/C]⋅[cos(30)i^ + sin(30)j^]
The significant surface in this part is side 4.
The area vector describing side 4 is...
ΦE = ∫E⋅dA
Φ4 = [500 N/C]⋅[cos(30)i^ + sin(30)j^]⋅(9×10-4 m2)j^
The area vector describing side 4 is...
→ A4 = (9×10-4 m2)j^
ΦE = ∫E⋅dA
Φ4 = [500 N/C]⋅[cos(30)i^ + sin(30)j^]⋅(9×10-4 m2)j^
→Φ4 = -0.23 N⋅m2/C
-----------------------------------------------------------------------------------------------------
PART E:
What is the net flux through these four sides?
Express your answer using two significant figures.
Express your answer using two significant figures.
SOLUTION:
Using the four fluxes determined from parts A through D,
Φnet = ∑Φ
Φnet = Φ1 + Φ2 + Φ3 + Φ4
Φnet = ∑Φ
Φnet = Φ1 + Φ2 + Φ3 + Φ4
→Φnet = 0 N⋅m2/C
Reason why the answer is negative in A and D has to do with the direction the field is in. Inward pointing vectors = negative flux, Outward pointing vectors = positive flux.
ReplyDeleteRiiiight. You are expected to summate the input and the output, so NET = out - in . If the field is entering the surface, and you only want to focus on that surface, NET = [out = 0] - [in = flux] --> therefore negative flux.
Delete